Which of the following is diamagnetic ?

Options

A. `(Ni(CN)_(4))^(2-)`
B. `(NiCl_(4))^(2-)`
C. `(CoCl_(4))^(2-)`
D. `(CoF_(6))^(2-)`

Explanation

To determine which of the following complexes is diamagnetic, we need to analyze the electronic configurations and the presence of unpaired electrons in each case. Let's break down the analysis step by step.

### Step 1: Identify the Complexes
We have two complexes to analyze:
1. Nickel cyanide tetra cyano complex (Ni(CN)₄)²⁻
2. Nickel chloride complex (NiCl₄)²⁻

### Step 2: Analyze the Nickel Cyanide Complex (Ni(CN)₄)²⁻
- **Oxidation State of Nickel**: In this complex, Nickel is in the +2 oxidation state.
- **Electronic Configuration**: The electronic configuration of neutral Nickel (Ni) is [Ar] 4s² 3d⁸. For Ni²⁺, it loses two electrons from the 4s orbital, resulting in [Ar] 3d⁸.
- **Ligand Type**: Cyanide (CN⁻) is a strong field ligand, which means it will cause pairing of electrons in the d-orbitals.
- **Electron Arrangement**: The 3d orbitals will be filled as follows:
- 3d: ↑↓ ↑↓ ↑↓ ↑↓ ↑ (8 electrons, all paired).
- **Hybridization**: The complex undergoes dsp² hybridization, resulting in a square planar geometry.
- **Unpaired Electrons**: There are no unpaired electrons in this complex.

### Conclusion for Nickel Cyanide Complex
Since there are no unpaired electrons, Ni(CN)₄²⁻ is **diamagnetic**.

### Step 3: Analyze the Nickel Chloride Complex (NiCl₄)²⁻
- **Oxidation State of Nickel**: In this complex, Nickel is also in the +2 oxidation state.
- **Electronic Configuration**: As before, Ni²⁺ has the configuration [Ar] 3d⁸.
- **Ligand Type**: Chloride (Cl⁻) is a weak field ligand, which does not cause pairing of electrons.
- **Electron Arrangement**: The 3d orbitals will be filled as follows:
- 3d: ↑ ↑ ↑ ↑ ↑ (8 electrons, with some unpaired).
- **Hybridization**: The complex undergoes sp³ hybridization, resulting in a tetrahedral geometry.
- **Unpaired Electrons**: There are 2 unpaired electrons in this complex.

### Conclusion for Nickel Chloride Complex
Since there are unpaired electrons, NiCl₄²⁻ is **paramagnetic**.

### Step 4: Final Comparison
After analyzing both complexes:
- Ni(CN)₄²⁻ is diamagnetic (no unpaired electrons).
- NiCl₄²⁻ is paramagnetic (two unpaired electrons).

### Final Answer
The diamagnetic complex is **Ni(CN)₄²⁻**.

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